jmc

Converting all of the given mixed numbers into fractions, we find $$\begin{array}{rl}3\frac{1}{5}& =3+\frac{1}{5}=\frac{3\cdot 5}{5}+\frac{1}{5}=\frac{15}{5}+\frac{1}{5}=\frac{16}{5},\\ 4\frac{1}{2}& =4+\frac{1}{2}=\frac{4\cdot 2}{2}+\frac{1}{2}=\frac{8}{2}+\frac{1}{2}=\frac{9}{2},\\ 2\frac{3}{4}& =2+\frac{3}{4}=\frac{2\cdot 4}{4}+\frac{3}{4}=\frac{8}{4}+\frac{3}{4}=\frac{11}{4}\\ 1\frac{2}{3}& =1+\frac{2}{3}=\frac{1\cdot 3}{3}+\frac{2}{3}=\frac{3}{3}+\frac{2}{3}=\frac{5}{3}.\end{array}$$ Substituting, we get $53\cdot \left(\frac{16}{5}-\frac{9}{2}\right)÷\left(\frac{11}{4}+\frac{5}{3}\right)$. We must first compute the expression inside parentheses. In order to add and subtract these fractions, we need to find a common denominator for the fractions. For the first set of parentheses, this is $5\cdot 2=10$ and for the second set it is $3\cdot 4=12$. Thus, we now have $$\begin{array}{rl}53\cdot \left(\frac{16}{5}-\frac{9}{2}\right)÷\left(\frac{11}{4}+\frac{5}{3}\right)& =53\cdot \left(\frac{16\cdot 2}{5\cdot 2}-\frac{9\cdot 5}{2\cdot 5}\right)\\ & ÷\left(\frac{11\cdot 3}{4\cdot 3}+\frac{5\cdot 4}{3\cdot 4}\right)\\ & =53\cdot \left(\frac{32}{10}-\frac{45}{10}\right)÷\left(\frac{33}{12}+\frac{20}{12}\right)\\ & =53\cdot \left(\frac{32-45}{10}\right)÷\left(\frac{33+20}{12}\right)\\ & =53\cdot \left(\frac{-13}{10}\right)÷\left(\frac{53}{12}\right)\\ & =53\cdot \left(\frac{-13}{10}\right)\cdot \left(\frac{12}{53}\right)\\ & =\cancel{53}\cdot \left(\frac{-13}{\text{cancelto}510}\right)\cdot \left(\frac{\text{cancelto}612}{\cancel{53}}\right)\\ & =\left(\frac{-13}{5}\right)\cdot \left(\frac{6}{1}\right)\\ & =\frac{(-13)\cdot (6)}{(5)\cdot (1)}\\ & =\frac{-78}{5}\\ & =-\frac{78}{5}.\end{array}$$ When we take 78 divided by 5, we get a quotient of 15 and a remainder of 3. In other words, $78=15\cdot 5+3$. Substituting into our fraction, $$\begin{array}{rl}-\frac{78}{5}& =-\frac{15\cdot 5+3}{5}\\ & =-\left(\frac{15\cdot 5}{5}+\frac{3}{5}\right)\\ & =-\left(\frac{15\cdot \cancel{5}}{\cancel{5}}+\frac{3}{5}\right)\\ & =-\left(15+\frac{3}{5}\right)\\ & =\boxed{-15\frac{3}{5}}.\end{array}$$